Return to Answer

Edit. While my first answer was more a suggestion on how to proceed, I decided to expand it in full. I also changed the former notation at some point in order to make it clearer, while still retaining the "duality bracket" notation for distributions instead of the "functional" one since using the latter would probably led to somewhat cumbersome formulas.

1. The first thing to note is that the generalization of the Radon transform to slowly increasing distributions and to fully general distributions uses the concept of adjoint (Radon) transform (see D. Ludwig, [1], or S. R. Deans, [2], chapter 5, pp. 109-124) (§5.4, pp. 120-130). First of all let's define the integral operator $\mathscr{R}^\dagger$ as $$\DeclareMathOperator{\dmu}{d\!} \mathscr{R}^\dagger \hat{\psi}(\mathbf{x})\triangleq \int\limits_{\lvert\xi\rvert=1} \hat{\psi}(\xi\cdot\mathbf{x},\xi)\dmu\xi\quad\mathbf{x}\in\Bbb R^{n}\label{1}\tag{AR} $$ where

   • $\hat{\psi}(p,\xi)=\mathscr{R}\psi (p,\xi)$ with $p\in\Bbb R$ is the Radon transform of a $C^\infty$ function $\psi$ and
   • $\xi\in\Bbb S^{n-1}=\big\{\xi\in\Bbb R^n:\lvert\xi\rvert=1\big\}$.
     

   This notation will be used throughout below: $\psi$ and likewise the all the (test) functions used below may be readily assumed to be rapidly decreasing, compactly supported, etc.: however, to keep the answer as elementary as possible, it is only assumed that their behavior at infinity is such that all the integration shown are meaningful (as it is the case for functions in $\mathscr{S}$ and $\mathscr{D}$). Then, for any two given and properly behaved $C^\infty$ functions $q$, $r$ we have that $$ \begin{split} \langle q, \mathscr{R}^\dagger \hat{r}\rangle & = \int\limits_{\Bbb R^n} q(\mathbf{x}){\mathscr{R}^\dagger\hat{r}}(\mathbf{x})\dmu\mathbf{x} \\ & = \int\limits_{\Bbb R^n} q(\mathbf{x})\bigg[\int\limits_{\lvert\xi\rvert=1}{\hat{r}}(\xi\cdot\mathbf{x},\xi)\dmu\xi\bigg]\dmu\mathbf{x}\\ & = \int\limits_{\Bbb R^n} q(\mathbf{x})\bigg[\int\limits_{\lvert\xi\rvert=1}\bigg(\int\limits_{-\infty}^{+\infty}{\hat{r}}(p,\xi)\delta(p-\xi\cdot\mathbf{x})\dmu p\bigg)\dmu\xi\bigg]\dmu\mathbf{x}\\ & = \int\limits_{\lvert\xi\rvert=1}\bigg[\int\limits_{-\infty}^{+\infty}{\hat{r}}(p,\xi)\bigg( \int\limits_{\Bbb R^n} q(\mathbf{x})\delta(p-\xi\cdot\mathbf{x})\dmu\mathbf{x}\bigg)\dmu p\bigg]\dmu\xi \\ & = \int\limits_{\lvert\xi\rvert=1}\,\,\int\limits_{-\infty}^{+\infty}{\hat{q}}(p,\xi)\hat{r}(p,\xi)\dmu p\dmu\xi =\langle \mathscr{R}{q},\hat{r}\rangle=\langle \hat{q},\hat{r}\rangle. \end{split}\label{2}\tag{GR} $$ Since the first member on the right of \eqref{2} can be readily interpreted as a lowly decreasing or general distribution, it is assumed as the definition of the Radon transform of a distribution, be it in $\mathscr{S}^\prime$ or in $\mathscr{D}^\prime$.

2. The second thing to note is that $\mathscr{R}^\dagger$ is verbatim the adjoint transform of the Radon transform, thus we have trivially that $$ \langle \mathscr{R}^\dagger \hat{q}, r\rangle\triangleq\langle \hat{q}, \mathscr{R} r \rangle=\langle \hat{q},\hat{r}\rangle. $$ This, as it will be shown in the next point, is important in order to understand the operational structure of $u$ as a distribution.

3. The third thing to note is that the function $u(\mathbf{x})$ in the OP we are trying to understand as a distribution is simply a translation of amplitude $a\in\Bbb R$ of the adjoint transform \eqref{1} of the Radon transform of some function. Precisely, still assuming that we are dealing with functions, we see that $$\DeclareMathOperator{\dmu}{d\!} u(\mathbf{x})= \int\limits_{\lvert\xi\rvert=1} {\alpha}(\xi\cdot\mathbf{x}+a,\xi)\dmu\xi = \mathscr{R}^\dagger {\alpha}_a(\mathbf{x}) \quad\mathbf{x}\in\Bbb R^{n}\label{3}\tag{AR'} $$ where we define $$ \alpha_a(p,\xi)\triangleq \alpha(p+a,\xi)\quad \forall a\in\Bbb R. $$

4. Finally, using the standard definitions of derivative and translation of a distribution (as for example given in [3]), we have all the elements in order to express $u$ as a general distribution: precisely. $$ \begin{split} {u}(\varphi) & = {\mathscr{R}^\dagger \alpha_a}(\varphi) = {\mathscr{R}^\dagger \left[\tfrac{\partial}{\partial p}\mathscr{R}g_a+ \tfrac{\partial^2}{\partial p^2}\mathscr{R}f_a\right]}(\varphi) \\ & = {\mathscr{R}^\dagger \left[\tfrac{\partial}{\partial p}\hat{g}_a+ \tfrac{\partial^2}{\partial p^2}\hat{f}_a\right]}(\varphi)\\ & = {\mathscr{R}^\dagger\tfrac{\partial}{\partial p}\hat{g}_a(\varphi) + \mathscr{R}^\dagger\tfrac{\partial^2}{\partial p^2}\hat{f}_a}(\varphi) \\ &= \Big\langle\tfrac{\partial}{\partial p}\hat{g}_a,\mathscr{R}\varphi\Big\rangle + \Big\langle\tfrac{\partial^2}{\partial p^2}\hat{f}_a,\mathscr{R}\varphi\Big\rangle \\ &= \Big\langle\tfrac{\partial}{\partial p}\hat{g}_a,\hat{\varphi}\Big\rangle + \Big\langle\tfrac{\partial^2}{\partial p^2}\hat{f}_a,\hat{\varphi}\Big\rangle\\ &= \Big\langle\tfrac{\partial}{\partial p}\hat{g},\hat{\varphi}_{-a}\Big\rangle + \Big\langle\tfrac{\partial^2}{\partial p^2}\hat{f},\hat{\varphi}_{-a}\Big\rangle\\ &= - \Big\langle\hat{g},\tfrac{\partial}{\partial p}\hat{\varphi}_{-a}\Big\rangle + \Big\langle\hat{f},\tfrac{\partial^2}{\partial p^2}\hat{\varphi}_{-a}\Big\rangle\\ &= - \Big\langle g, \mathscr{R}^\dagger\tfrac{\partial}{\partial p}\hat{\varphi}_{-a}\Big\rangle + \Big\langle f,\mathscr{R}^\dagger\tfrac{\partial^2}{\partial p^2}\hat{\varphi}_{-a}\Big\rangle \end{split} $$

References

[1] Stanley Roderick Deans, The Radon transform and some of its applications, revised reprint of the 1983 original edition (English), Malabar, FL: Publishing Co., ISBN 0-89464-718-0, pp. xii+295 (1993), MR1274701, Zbl 0868.44001.

[2] Donald Ludwig, "The Radon transform on Euclidean space", (English) Communication in Pure and Applied Mathematics 19, 49-81 (1966), MR0190652, Zbl 0134.11305.

[3] Vasily Sergeevich Vladimirov (2002), Methods of the theory of generalized functions, Analytical Methods and Special Functions, Vol. 6, London–New York: Taylor & Francis, pp. XII+353, ISBN 0-415-27356-0, MR2012831, Zbl 1078.46029.

Edit. While my first answer was more a suggestion on how to proceed, I decided to expand it in full. I also changed the former notation at some point in order to make it clearer, while still retaining the "duality bracket" notation for distributions instead of the "functional" one since using the latter would probably led to somewhat cumbersome formulas.

1. The first thing to note is that the generalization of the Radon transform to slowly increasing distributions and to fully general distributions uses the concept of adjoint (Radon) transform (see D. Ludwig, [1], or S. R. Deans, [2], chapter 5, pp. 109-124) (§5.4, pp. 120-130). First of all let's define the integral operator $\mathscr{R}^\dagger$ as $$\DeclareMathOperator{\dmu}{d\!} \mathscr{R}^\dagger \hat{\psi}(\mathbf{x})\triangleq \int\limits_{\lvert\xi\rvert=1} \hat{\psi}(\xi\cdot\mathbf{x},\xi)\dmu\xi\quad\mathbf{x}\in\Bbb R^{n}\label{1}\tag{AR} $$ where

   • $\hat{\psi}(p,\xi)=\mathscr{R}\psi (p,\xi)$ with $p\in\Bbb R$ is the Radon transform of a $C^\infty$ function $\psi$ and
   • $\xi\in\Bbb S^{n-1}=\big\{\xi\in\Bbb R^n:\lvert\xi\rvert=1\big\}$.
     

   This notation will be used throughout below: $\psi$ and likewise the all the (test) functions used below may be readily assumed to be rapidly decreasing, compactly supported, etc.: however, to keep the answer as elementary as possible, it is only assumed that their behavior at infinity is such that all the integration shown are meaningful (as it is the case for functions in $\mathscr{S}$ and $\mathscr{D}$). Then, for any two given and properly behaved $C^\infty$ functions $q$, $r$ we have that $$ \begin{split} \langle q, \mathscr{R}^\dagger \hat{r}\rangle & = \int\limits_{\Bbb R^n} q(\mathbf{x}){\mathscr{R}^\dagger\hat{r}}(\mathbf{x})\dmu\mathbf{x} \\ & = \int\limits_{\Bbb R^n} q(\mathbf{x})\bigg[\int\limits_{\lvert\xi\rvert=1}{\hat{r}}(\xi\cdot\mathbf{x},\xi)\dmu\xi\bigg]\dmu\mathbf{x}\\ & = \int\limits_{\Bbb R^n} q(\mathbf{x})\bigg[\int\limits_{\lvert\xi\rvert=1}\bigg(\int\limits_{-\infty}^{+\infty}{\hat{r}}(p,\xi)\delta(p-\xi\cdot\mathbf{x})\dmu p\bigg)\dmu\xi\bigg]\dmu\mathbf{x}\\ & = \int\limits_{\lvert\xi\rvert=1}\bigg[\int\limits_{-\infty}^{+\infty}{\hat{r}}(p,\xi)\bigg( \int\limits_{\Bbb R^n} q(\mathbf{x})\delta(p-\xi\cdot\mathbf{x})\dmu\mathbf{x}\bigg)\dmu p\bigg]\dmu\xi \\ & = \int\limits_{\lvert\xi\rvert=1}\,\,\int\limits_{-\infty}^{+\infty}{\hat{q}}(p,\xi)\hat{r}(p,\xi)\dmu p\dmu\xi =\langle \mathscr{R}{q},\hat{r}\rangle=\langle \hat{q},\hat{r}\rangle. \end{split}\label{2}\tag{GR} $$ Since the first member on the right of \eqref{2} can be readily interpreted as a slowly decreasing or general distribution, it is assumed as the definition of the Radon transform of a distribution, be it in $\mathscr{S}^\prime$ or in $\mathscr{D}^\prime$.

2. The second thing to note is that $\mathscr{R}^\dagger$ is verbatim the adjoint transform of the Radon transform, thus we have trivially that $$ \langle \mathscr{R}^\dagger \hat{q}, r\rangle\triangleq\langle \hat{q}, \mathscr{R} r \rangle=\langle \hat{q},\hat{r}\rangle. $$ This, as it will be shown in the next point, is important in order to understand the operational structure of $u$ as a distribution.

3. The third thing to note is that the function $u(\mathbf{x})$ in the OP we are trying to understand as a distribution is simply a translation of amplitude $a\in\Bbb R$ of the adjoint transform \eqref{1} of the Radon transform of some function. Precisely, still assuming that we are dealing with functions, we see that $$\DeclareMathOperator{\dmu}{d\!} u(\mathbf{x})= \int\limits_{\lvert\xi\rvert=1} {\alpha}(\xi\cdot\mathbf{x}+a,\xi)\dmu\xi = \mathscr{R}^\dagger {\alpha}_a(\mathbf{x}) \quad\mathbf{x}\in\Bbb R^{n}\label{3}\tag{AR'} $$ where we define $$ \alpha_a(p,\xi)\triangleq \alpha(p+a,\xi)\quad \forall a\in\Bbb R. $$

4. Finally, using the standard definitions of derivative and translation of a distribution (as for example given in [3]), we have all the elements in order to express $u$ as a general distribution: precisely. $$ \begin{split} {u}(\varphi) & = {\mathscr{R}^\dagger \alpha_a}(\varphi) = {\mathscr{R}^\dagger \left[\tfrac{\partial}{\partial p}\mathscr{R}g_a+ \tfrac{\partial^2}{\partial p^2}\mathscr{R}f_a\right]}(\varphi) \\ & = {\mathscr{R}^\dagger \left[\tfrac{\partial}{\partial p}\hat{g}_a+ \tfrac{\partial^2}{\partial p^2}\hat{f}_a\right]}(\varphi)\\ & = {\mathscr{R}^\dagger\tfrac{\partial}{\partial p}\hat{g}_a(\varphi) + \mathscr{R}^\dagger\tfrac{\partial^2}{\partial p^2}\hat{f}_a}(\varphi) \\ &= \Big\langle\tfrac{\partial}{\partial p}\hat{g}_a,\mathscr{R}\varphi\Big\rangle + \Big\langle\tfrac{\partial^2}{\partial p^2}\hat{f}_a,\mathscr{R}\varphi\Big\rangle \\ &= \Big\langle\tfrac{\partial}{\partial p}\hat{g}_a,\hat{\varphi}\Big\rangle + \Big\langle\tfrac{\partial^2}{\partial p^2}\hat{f}_a,\hat{\varphi}\Big\rangle\\ &= \Big\langle\tfrac{\partial}{\partial p}\hat{g},\hat{\varphi}_{-a}\Big\rangle + \Big\langle\tfrac{\partial^2}{\partial p^2}\hat{f},\hat{\varphi}_{-a}\Big\rangle\\ &= - \Big\langle\hat{g},\tfrac{\partial}{\partial p}\hat{\varphi}_{-a}\Big\rangle + \Big\langle\hat{f},\tfrac{\partial^2}{\partial p^2}\hat{\varphi}_{-a}\Big\rangle\\ &= - \Big\langle g, \mathscr{R}^\dagger\tfrac{\partial}{\partial p}\hat{\varphi}_{-a}\Big\rangle + \Big\langle f,\mathscr{R}^\dagger\tfrac{\partial^2}{\partial p^2}\hat{\varphi}_{-a}\Big\rangle \end{split} $$

References

[1] Stanley Roderick Deans, The Radon transform and some of its applications, revised reprint of the 1983 original edition (English), Malabar, FL: Publishing Co., ISBN 0-89464-718-0, pp. xii+295 (1993), MR1274701, Zbl 0868.44001.

[2] Donald Ludwig, "The Radon transform on Euclidean space", (English) Communication in Pure and Applied Mathematics 19, 49-81 (1966), MR0190652, Zbl 0134.11305.

[3] Vasily Sergeevich Vladimirov (2002), Methods of the theory of generalized functions, Analytical Methods and Special Functions, Vol. 6, London–New York: Taylor & Francis, pp. XII+353, ISBN 0-415-27356-0, MR2012831, Zbl 1078.46029.

Edit. While my first answer was more a suggestion on how to proceed, I decided to expand it in full. I also changed the former notation at some point in order to make it clearer, while still retaining the "duality bracket" notation for distributions instead of the "functional" since using the latter would probably led to somewhat more cumbersome formulas.

1. The first thing to note is that the generalization of the Radon transform to slowly increasing distributions and to fully general distributions uses the concept of adjoint (Radon) transform (see D. Ludwig, [1], or S. R. Deans, [2], chapter 5, pp. 109-124) (§5.4, pp. 120-130). First of all let's define the integral operator $\mathscr{R}^\dagger$ as $$\DeclareMathOperator{\dmu}{d\!} \mathscr{R}^\dagger \hat{\psi}(\mathbf{x})\triangleq \int\limits_{\lvert\xi\rvert=1} \hat{\psi}(\xi\cdot\mathbf{x},\xi)\dmu\xi\quad\mathbf{x}\in\Bbb R^{n}\label{1}\tag{AR} $$ where

   • $\hat{\psi}(p,\xi)=\mathscr{R}\psi (p,\xi)$ with $p\in\Bbb R$ is the Radon transform of a $C^\infty$ function $\psi$ and
   • $\xi\in\Bbb S^{n-1}=\big\{\xi\in\Bbb R^n:\lvert\xi\rvert=1\big\}$.
     

   This notation will be used throughout below: $\psi$ and likewise the all the (test) functions used below may be readily assumed to be rapidly decreasing, compactly supported, etc.: however, to keep the answer as elementary as possible, it is only assumed that their behavior at infinity is such that all the integration shown are meaningful (as it is the case for functions in $\mathscr{S}$ and $\mathscr{D}$). Then, for any two given and properly behaved $C^\infty$ functions $q$, $r$ we have that $$ \begin{split} \langle q, \mathscr{R}^\dagger \hat{r}\rangle & = \int\limits_{\Bbb R^n} q(\mathbf{x}){\mathscr{R}^\dagger\hat{r}}(\mathbf{x})\dmu\mathbf{x} \\ & = \int\limits_{\Bbb R^n} q(\mathbf{x})\bigg[\int\limits_{\lvert\xi\rvert=1}{\hat{r}}(\xi\cdot\mathbf{x},\xi)\dmu\xi\bigg]\dmu\mathbf{x}\\ & = \int\limits_{\Bbb R^n} q(\mathbf{x})\bigg[\int\limits_{\lvert\xi\rvert=1}\bigg(\int\limits_{-\infty}^{+\infty}{\hat{r}}(p,\xi)\delta(p-\xi\cdot\mathbf{x})\dmu p\bigg)\dmu\xi\bigg]\dmu\mathbf{x}\\ & = \int\limits_{\lvert\xi\rvert=1}\bigg[\int\limits_{-\infty}^{+\infty}{\hat{r}}(p,\xi)\bigg( \int\limits_{\Bbb R^n} q(\mathbf{x})\delta(p-\xi\cdot\mathbf{x})\dmu\mathbf{x}\bigg)\dmu p\bigg]\dmu\xi \\ & = \int\limits_{\lvert\xi\rvert=1}\,\,\int\limits_{-\infty}^{+\infty}{\hat{q}}(p,\xi)\hat{r}(p,\xi)\dmu p\dmu\xi =\langle \mathscr{R}{q},\hat{r}\rangle=\langle \hat{q},\hat{r}\rangle. \end{split}\label{2}\tag{GR} $$ Since the first member on the right of \eqref{2} can be readily interpreted as a lowly decreasing or general distribution, it is assumed as the definition of the Radon transform of a distribution, be it in $\mathscr{S}^\prime$ or in $\mathscr{D}^\prime$.

2. The second thing to note is that $\mathscr{R}^\dagger$ is verbatim the adjoint transform of the Radon transform, thus we have trivially that $$ \langle \mathscr{R}^\dagger \hat{q}, r\rangle\triangleq\langle \hat{q}, \mathscr{R} r \rangle=\langle \hat{q},\hat{r}\rangle. $$ This, as it will be shown in the next point, is important in order to understand the operational structure of $u$ as a distribution.

3. The third thing to note is that the function $u(\mathbf{x})$ in the OP we are trying to understand as a distribution is simply a translation of amplitude $a\in\Bbb R$ of the adjoint transform \eqref{1} of the Radon transform of some function. Precisely, still assuming that we are dealing with functions, we see that $$\DeclareMathOperator{\dmu}{d\!} u(\mathbf{x})= \int\limits_{\lvert\xi\rvert=1} {\alpha}(\xi\cdot\mathbf{x}+a,\xi)\dmu\xi = \mathscr{R}^\dagger {\alpha}_a(\mathbf{x}) \quad\mathbf{x}\in\Bbb R^{n}\label{3}\tag{AR'} $$ where we define $$ \alpha_a(p,\xi)\triangleq \alpha(p+a,\xi)\quad \forall a\in\Bbb R. $$

4. Finally, using the standard definitions of derivative and translation of a distribution (as for example given in [3]), we have all the elements in order to express $u$ as a general distribution: precisely. $$ \begin{split} {u}(\varphi) & = {\mathscr{R}^\dagger \alpha_a}(\varphi) = {\mathscr{R}^\dagger \left[\tfrac{\partial}{\partial p}\mathscr{R}g_a+ \tfrac{\partial^2}{\partial p^2}\mathscr{R}f_a\right]}(\varphi) \\ & = {\mathscr{R}^\dagger \left[\tfrac{\partial}{\partial p}\hat{g}_a+ \tfrac{\partial^2}{\partial p^2}\hat{f}_a\right]}(\varphi)\\ & = {\mathscr{R}^\dagger\tfrac{\partial}{\partial p}\hat{g}_a(\varphi) + \mathscr{R}^\dagger\tfrac{\partial^2}{\partial p^2}\hat{f}_a}(\varphi) \\ &= \Big\langle\tfrac{\partial}{\partial p}\hat{g}_a,\mathscr{R}\varphi\Big\rangle + \Big\langle\tfrac{\partial^2}{\partial p^2}\hat{f}_a,\mathscr{R}\varphi\Big\rangle \\ &= \Big\langle\tfrac{\partial}{\partial p}\hat{g}_a,\hat{\varphi}\Big\rangle + \Big\langle\tfrac{\partial^2}{\partial p^2}\hat{f}_a,\hat{\varphi}\Big\rangle\\ &= \Big\langle\tfrac{\partial}{\partial p}\hat{g},\hat{\varphi}_{-a}\Big\rangle + \Big\langle\tfrac{\partial^2}{\partial p^2}\hat{f},\hat{\varphi}_{-a}\Big\rangle\\ &= - \Big\langle\hat{g},\tfrac{\partial}{\partial p}\hat{\varphi}_{-a}\Big\rangle + \Big\langle\hat{f},\tfrac{\partial^2}{\partial p^2}\hat{\varphi}_{-a}\Big\rangle\\ &= - \Big\langle g, \mathscr{R}^\dagger\tfrac{\partial}{\partial p}\hat{\varphi}_{-a}\Big\rangle + \Big\langle f,\mathscr{R}^\dagger\tfrac{\partial^2}{\partial p^2}\hat{\varphi}_{-a}\Big\rangle \end{split} $$

References

[1] Stanley Roderick Deans, The Radon transform and some of its applications, revised reprint of the 1983 original edition (English), Malabar, FL: Publishing Co., ISBN 0-89464-718-0, pp. xii+295 (1993), MR1274701, Zbl 0868.44001.

[2] Donald Ludwig, "The Radon transform on Euclidean space", (English) Communication in Pure and Applied Mathematics 19, 49-81 (1966), MR0190652, Zbl 0134.11305.

[3] Vasily Sergeevich Vladimirov (2002), Methods of the theory of generalized functions, Analytical Methods and Special Functions, Vol. 6, London–New York: Taylor & Francis, pp. XII+353, ISBN 0-415-27356-0, MR2012831, Zbl 1078.46029.

Edit. While my first answer was more a suggestion on how to proceed, I decided to expand it in full. I also changed the former notation at some point in order to make it clearer, while still retaining the "duality bracket" notation for distributions instead of the "functional" one since using the latter would probably led to somewhat cumbersome formulas.

1. The first thing to note is that the generalization of the Radon transform to slowly increasing distributions and to fully general distributions uses the concept of adjoint (Radon) transform (see D. Ludwig, [1], or S. R. Deans, [2], chapter 5, pp. 109-124) (§5.4, pp. 120-130). First of all let's define the integral operator $\mathscr{R}^\dagger$ as $$\DeclareMathOperator{\dmu}{d\!} \mathscr{R}^\dagger \hat{\psi}(\mathbf{x})\triangleq \int\limits_{\lvert\xi\rvert=1} \hat{\psi}(\xi\cdot\mathbf{x},\xi)\dmu\xi\quad\mathbf{x}\in\Bbb R^{n}\label{1}\tag{AR} $$ where

   • $\hat{\psi}(p,\xi)=\mathscr{R}\psi (p,\xi)$ with $p\in\Bbb R$ is the Radon transform of a $C^\infty$ function $\psi$ and
   • $\xi\in\Bbb S^{n-1}=\big\{\xi\in\Bbb R^n:\lvert\xi\rvert=1\big\}$.
     

   This notation will be used throughout below: $\psi$ and likewise the all the (test) functions used below may be readily assumed to be rapidly decreasing, compactly supported, etc.: however, to keep the answer as elementary as possible, it is only assumed that their behavior at infinity is such that all the integration shown are meaningful (as it is the case for functions in $\mathscr{S}$ and $\mathscr{D}$). Then, for any two given and properly behaved $C^\infty$ functions $q$, $r$ we have that $$ \begin{split} \langle q, \mathscr{R}^\dagger \hat{r}\rangle & = \int\limits_{\Bbb R^n} q(\mathbf{x}){\mathscr{R}^\dagger\hat{r}}(\mathbf{x})\dmu\mathbf{x} \\ & = \int\limits_{\Bbb R^n} q(\mathbf{x})\bigg[\int\limits_{\lvert\xi\rvert=1}{\hat{r}}(\xi\cdot\mathbf{x},\xi)\dmu\xi\bigg]\dmu\mathbf{x}\\ & = \int\limits_{\Bbb R^n} q(\mathbf{x})\bigg[\int\limits_{\lvert\xi\rvert=1}\bigg(\int\limits_{-\infty}^{+\infty}{\hat{r}}(p,\xi)\delta(p-\xi\cdot\mathbf{x})\dmu p\bigg)\dmu\xi\bigg]\dmu\mathbf{x}\\ & = \int\limits_{\lvert\xi\rvert=1}\bigg[\int\limits_{-\infty}^{+\infty}{\hat{r}}(p,\xi)\bigg( \int\limits_{\Bbb R^n} q(\mathbf{x})\delta(p-\xi\cdot\mathbf{x})\dmu\mathbf{x}\bigg)\dmu p\bigg]\dmu\xi \\ & = \int\limits_{\lvert\xi\rvert=1}\,\,\int\limits_{-\infty}^{+\infty}{\hat{q}}(p,\xi)\hat{r}(p,\xi)\dmu p\dmu\xi =\langle \mathscr{R}{q},\hat{r}\rangle=\langle \hat{q},\hat{r}\rangle. \end{split}\label{2}\tag{GR} $$ Since the first member on the right of \eqref{2} can be readily interpreted as a lowly decreasing or general distribution, it is assumed as the definition of the Radon transform of a distribution, be it in $\mathscr{S}^\prime$ or in $\mathscr{D}^\prime$.

2. The second thing to note is that $\mathscr{R}^\dagger$ is verbatim the adjoint transform of the Radon transform, thus we have trivially that $$ \langle \mathscr{R}^\dagger \hat{q}, r\rangle\triangleq\langle \hat{q}, \mathscr{R} r \rangle=\langle \hat{q},\hat{r}\rangle. $$ This, as it will be shown in the next point, is important in order to understand the operational structure of $u$ as a distribution.

3. The third thing to note is that the function $u(\mathbf{x})$ in the OP we are trying to understand as a distribution is simply a translation of amplitude $a\in\Bbb R$ of the adjoint transform \eqref{1} of the Radon transform of some function. Precisely, still assuming that we are dealing with functions, we see that $$\DeclareMathOperator{\dmu}{d\!} u(\mathbf{x})= \int\limits_{\lvert\xi\rvert=1} {\alpha}(\xi\cdot\mathbf{x}+a,\xi)\dmu\xi = \mathscr{R}^\dagger {\alpha}_a(\mathbf{x}) \quad\mathbf{x}\in\Bbb R^{n}\label{3}\tag{AR'} $$ where we define $$ \alpha_a(p,\xi)\triangleq \alpha(p+a,\xi)\quad \forall a\in\Bbb R. $$

4. Finally, using the standard definitions of derivative and translation of a distribution (as for example given in [3]), we have all the elements in order to express $u$ as a general distribution: precisely. $$ \begin{split} {u}(\varphi) & = {\mathscr{R}^\dagger \alpha_a}(\varphi) = {\mathscr{R}^\dagger \left[\tfrac{\partial}{\partial p}\mathscr{R}g_a+ \tfrac{\partial^2}{\partial p^2}\mathscr{R}f_a\right]}(\varphi) \\ & = {\mathscr{R}^\dagger \left[\tfrac{\partial}{\partial p}\hat{g}_a+ \tfrac{\partial^2}{\partial p^2}\hat{f}_a\right]}(\varphi)\\ & = {\mathscr{R}^\dagger\tfrac{\partial}{\partial p}\hat{g}_a(\varphi) + \mathscr{R}^\dagger\tfrac{\partial^2}{\partial p^2}\hat{f}_a}(\varphi) \\ &= \Big\langle\tfrac{\partial}{\partial p}\hat{g}_a,\mathscr{R}\varphi\Big\rangle + \Big\langle\tfrac{\partial^2}{\partial p^2}\hat{f}_a,\mathscr{R}\varphi\Big\rangle \\ &= \Big\langle\tfrac{\partial}{\partial p}\hat{g}_a,\hat{\varphi}\Big\rangle + \Big\langle\tfrac{\partial^2}{\partial p^2}\hat{f}_a,\hat{\varphi}\Big\rangle\\ &= \Big\langle\tfrac{\partial}{\partial p}\hat{g},\hat{\varphi}_{-a}\Big\rangle + \Big\langle\tfrac{\partial^2}{\partial p^2}\hat{f},\hat{\varphi}_{-a}\Big\rangle\\ &= - \Big\langle\hat{g},\tfrac{\partial}{\partial p}\hat{\varphi}_{-a}\Big\rangle + \Big\langle\hat{f},\tfrac{\partial^2}{\partial p^2}\hat{\varphi}_{-a}\Big\rangle\\ &= - \Big\langle g, \mathscr{R}^\dagger\tfrac{\partial}{\partial p}\hat{\varphi}_{-a}\Big\rangle + \Big\langle f,\mathscr{R}^\dagger\tfrac{\partial^2}{\partial p^2}\hat{\varphi}_{-a}\Big\rangle \end{split} $$

References

[1] Stanley Roderick Deans, The Radon transform and some of its applications, revised reprint of the 1983 original edition (English), Malabar, FL: Publishing Co., ISBN 0-89464-718-0, pp. xii+295 (1993), MR1274701, Zbl 0868.44001.

[2] Donald Ludwig, "The Radon transform on Euclidean space", (English) Communication in Pure and Applied Mathematics 19, 49-81 (1966), MR0190652, Zbl 0134.11305.

[3] Vasily Sergeevich Vladimirov (2002), Methods of the theory of generalized functions, Analytical Methods and Special Functions, Vol. 6, London–New York: Taylor & Francis, pp. XII+353, ISBN 0-415-27356-0, MR2012831, Zbl 1078.46029.

Edit. While my first answer was more a suggestion on how to proceed, I decided to expand it in full. I also changed the former notation at some point in order to make it clearer, while still retaining the "duality bracket" notation for distributions instead of the "functional" since using the latter would probably led to somewhat more cumbersome formulas.

1. The first thing to note is that the generalization of the Radon transform to slowly increasing distributions and to fully general distributions uses the concept of adjoint (Radon) transform (see D. Ludwig, [1], or S. R. Deans, [2], chapter 5, pp. 109-124) (§5.4, pp. 120-130). First of all let's define the integral operator $\mathscr{R}^\dagger$ as $$\DeclareMathOperator{\dmu}{d\!} \mathscr{R}^\dagger \hat{\psi}(\mathbf{x})\triangleq \int\limits_{\lvert\xi\rvert=1} \hat{\psi}(\xi\cdot\mathbf{x},\xi)\dmu\xi\quad\mathbf{x}\in\Bbb R^{n}\label{1}\tag{AR} $$ where

   • $\hat{\psi}(p,\xi)=\mathscr{R}\psi (p,\xi)$ with $p\in\Bbb R$ is the Radon transform of a $C^\infty$ function $\psi$ and
   • $\xi\in\Bbb S^{n-1}=\big\{\xi\in\Bbb R^n:\lvert\xi\rvert=1\big\}$.
     

   This notation will be used throughout below: $\psi$ and likewise the all the (test) functions used below may be readily assumed to be rapidly decreasing, compactly supported, etc.: however, to keep the answer as elementary as possible, it is only assumed that their behavior at infinity is such that all the integration shown are meaningful (as it is the case for functions in $\mathscr{S}$ and $\mathscr{D}$). Then, for any two given and properly behaved $C^\infty$ functions $q$, $r$ we have that $$ \begin{split} \langle q, \mathscr{R}^\dagger \hat{r}\rangle & = \int\limits_{\Bbb R^n} q(\mathbf{x}){\mathscr{R}^\dagger\hat{r}}(\mathbf{x})\dmu\mathbf{x} \\ & = \int\limits_{\Bbb R^n} q(\mathbf{x})\bigg[\int\limits_{\lvert\xi\rvert=1}{\hat{r}}(\xi\cdot\mathbf{x},\xi)\dmu\xi\bigg]\dmu\mathbf{x}\\ & = \int\limits_{\Bbb R^n} q(\mathbf{x})\bigg[\int\limits_{\lvert\xi\rvert=1}\bigg(\int\limits_{-\infty}^{+\infty}{\hat{r}}(p,\xi)\delta(p-\xi\cdot\mathbf{x})\dmu p\bigg)\dmu\xi\bigg]\dmu\mathbf{x}\\ & = \int\limits_{\lvert\xi\rvert=1}\bigg[\int\limits_{-\infty}^{+\infty}{\hat{r}}(p,\xi)\bigg( \int\limits_{\Bbb R^n} q(\mathbf{x})\delta(p-\xi\cdot\mathbf{x})\dmu\mathbf{x}\bigg)\dmu p\bigg]\dmu\xi \\ & = \int\limits_{\lvert\xi\rvert=1}\,\,\int\limits_{-\infty}^{+\infty}{\hat{q}}(p,\xi)\hat{r}(p,\xi)\dmu p\dmu\xi =\langle \mathscr{R}{q},\hat{r}\rangle=\langle \hat{q},\hat{r}\rangle. \end{split}\label{2}\tag{GR} $$ Since the first member on the right of \eqref{2} can be readily interpreted as a lowly decreasing or general distribution, it is assumed as the definition of the Radon transform of a distribution, be it in $\mathscr{S}^\prime$ or in $\mathscr{D}^\prime$.

2. The second thing to note is that $\mathscr{R}^\dagger$ is verbatim the adjoint transform of the Radon transform, thus we have trivially that $$ \langle \mathscr{R}^\dagger \hat{q}, r\rangle\triangleq\langle \hat{q}, \mathscr{R} r \rangle=\langle \hat{q},\hat{r}\rangle. $$ This, as it will be shown in the next point, is important in order to understand the operational structure of $u$ as a distribution.

3. The third thing to note is that the function $u(\mathbf{x})$ in the OP we are trying to understand as a distribution is simply a translation of amplitude $a\in\Bbb R$ of the adjoint transform \eqref{1} of the Radon transform of some function. Precisely, still assuming that we are dealing with functions, we see that $$\DeclareMathOperator{\dmu}{d\!} u(\mathbf{x})= \int\limits_{\lvert\xi\rvert=1} {\alpha}(\xi\cdot\mathbf{x}+a,\xi)\dmu\xi = \mathscr{R}^\dagger {\alpha}_a(\mathbf{x}) \quad\mathbf{x}\in\Bbb R^{n}\label{3}\tag{AR'} $$ where we define $$ \alpha_a(p,\xi)\triangleq \alpha(p+a,\xi)\quad \forall a\in\Bbb R. $$

4. Finally, using the standard definitions of derivative and translation of a distribution (as for example given in [3]), we have all the elements in order to express $u$ as a general distribution: precisely. $$ \begin{split} {u}(\varphi) & = {\mathscr{R}^\dagger \alpha_a}(\varphi) = {\mathscr{R}^\dagger \left[\tfrac{\partial}{\partial p}\mathscr{R}g_a+ \tfrac{\partial^2}{\partial p^2}\mathscr{R}f_a\right]}(\varphi) \\ & = {\mathscr{R}^\dagger \left[\tfrac{\partial}{\partial p}\hat{g}_a+ \tfrac{\partial^2}{\partial p^2}\hat{f}_a\right]}(\varphi)\\ & = {\mathscr{R}^\dagger\tfrac{\partial}{\partial p}\hat{g}_a(\varphi) + \mathscr{R}^\dagger\tfrac{\partial^2}{\partial p^2}\hat{f}_a}(\varphi) \\ &= - \left\langle g, \mathscr{R}^\dagger\tfrac{\partial}{\partial p}\hat{\varphi}_{-a}\right\rangle + \left\langle f,\mathscr{R}^\dagger\tfrac{\partial^2}{\partial p^2}\hat{\varphi}_{-a}\right\rangle \end{split} $$

References

[1] Stanley Roderick Deans, The Radon transform and some of its applications, revised reprint of the 1983 original edition (English), Malabar, FL: Publishing Co., ISBN 0-89464-718-0, pp. xii+295 (1993), MR1274701, Zbl 0868.44001.

[2] Donald Ludwig, "The Radon transform on Euclidean space", (English) Communication in Pure and Applied Mathematics 19, 49-81 (1966), MR0190652, Zbl 0134.11305.

[3] Vasily Sergeevich Vladimirov (2002), Methods of the theory of generalized functions, Analytical Methods and Special Functions, Vol. 6, London–New York: Taylor & Francis, pp. XII+353, ISBN 0-415-27356-0, MR2012831, Zbl 1078.46029.

Edit. While my first answer was more a suggestion on how to proceed, I decided to expand it in full. I also changed the former notation at some point in order to make it clearer, while still retaining the "duality bracket" notation for distributions instead of the "functional" since using the latter would probably led to somewhat more cumbersome formulas.

1. The first thing to note is that the generalization of the Radon transform to slowly increasing distributions and to fully general distributions uses the concept of adjoint (Radon) transform (see D. Ludwig, [1], or S. R. Deans, [2], chapter 5, pp. 109-124) (§5.4, pp. 120-130). First of all let's define the integral operator $\mathscr{R}^\dagger$ as $$\DeclareMathOperator{\dmu}{d\!} \mathscr{R}^\dagger \hat{\psi}(\mathbf{x})\triangleq \int\limits_{\lvert\xi\rvert=1} \hat{\psi}(\xi\cdot\mathbf{x},\xi)\dmu\xi\quad\mathbf{x}\in\Bbb R^{n}\label{1}\tag{AR} $$ where

   • $\hat{\psi}(p,\xi)=\mathscr{R}\psi (p,\xi)$ with $p\in\Bbb R$ is the Radon transform of a $C^\infty$ function $\psi$ and
   • $\xi\in\Bbb S^{n-1}=\big\{\xi\in\Bbb R^n:\lvert\xi\rvert=1\big\}$.
     

   This notation will be used throughout below: $\psi$ and likewise the all the (test) functions used below may be readily assumed to be rapidly decreasing, compactly supported, etc.: however, to keep the answer as elementary as possible, it is only assumed that their behavior at infinity is such that all the integration shown are meaningful (as it is the case for functions in $\mathscr{S}$ and $\mathscr{D}$). Then, for any two given and properly behaved $C^\infty$ functions $q$, $r$ we have that $$ \begin{split} \langle q, \mathscr{R}^\dagger \hat{r}\rangle & = \int\limits_{\Bbb R^n} q(\mathbf{x}){\mathscr{R}^\dagger\hat{r}}(\mathbf{x})\dmu\mathbf{x} \\ & = \int\limits_{\Bbb R^n} q(\mathbf{x})\bigg[\int\limits_{\lvert\xi\rvert=1}{\hat{r}}(\xi\cdot\mathbf{x},\xi)\dmu\xi\bigg]\dmu\mathbf{x}\\ & = \int\limits_{\Bbb R^n} q(\mathbf{x})\bigg[\int\limits_{\lvert\xi\rvert=1}\bigg(\int\limits_{-\infty}^{+\infty}{\hat{r}}(p,\xi)\delta(p-\xi\cdot\mathbf{x})\dmu p\bigg)\dmu\xi\bigg]\dmu\mathbf{x}\\ & = \int\limits_{\lvert\xi\rvert=1}\bigg[\int\limits_{-\infty}^{+\infty}{\hat{r}}(p,\xi)\bigg( \int\limits_{\Bbb R^n} q(\mathbf{x})\delta(p-\xi\cdot\mathbf{x})\dmu\mathbf{x}\bigg)\dmu p\bigg]\dmu\xi \\ & = \int\limits_{\lvert\xi\rvert=1}\,\,\int\limits_{-\infty}^{+\infty}{\hat{q}}(p,\xi)\hat{r}(p,\xi)\dmu p\dmu\xi =\langle \mathscr{R}{q},\hat{r}\rangle=\langle \hat{q},\hat{r}\rangle. \end{split}\label{2}\tag{GR} $$ Since the first member on the right of \eqref{2} can be readily interpreted as a lowly decreasing or general distribution, it is assumed as the definition of the Radon transform of a distribution, be it in $\mathscr{S}^\prime$ or in $\mathscr{D}^\prime$.

2. The second thing to note is that $\mathscr{R}^\dagger$ is verbatim the adjoint transform of the Radon transform, thus we have trivially that $$ \langle \mathscr{R}^\dagger \hat{q}, r\rangle\triangleq\langle \hat{q}, \mathscr{R} r \rangle=\langle \hat{q},\hat{r}\rangle. $$ This, as it will be shown in the next point, is important in order to understand the operational structure of $u$ as a distribution.

3. The third thing to note is that the function $u(\mathbf{x})$ in the OP we are trying to understand as a distribution is simply a translation of amplitude $a\in\Bbb R$ of the adjoint transform \eqref{1} of the Radon transform of some function. Precisely, still assuming that we are dealing with functions, we see that $$\DeclareMathOperator{\dmu}{d\!} u(\mathbf{x})= \int\limits_{\lvert\xi\rvert=1} {\alpha}(\xi\cdot\mathbf{x}+a,\xi)\dmu\xi = \mathscr{R}^\dagger {\alpha}_a(\mathbf{x}) \quad\mathbf{x}\in\Bbb R^{n}\label{3}\tag{AR'} $$ where we define $$ \alpha_a(p,\xi)\triangleq \alpha(p+a,\xi)\quad \forall a\in\Bbb R. $$

4. Finally, using the standard definitions of derivative and translation of a distribution (as for example given in [3]), we have all the elements in order to express $u$ as a general distribution: precisely. $$ \begin{split} {u}(\varphi) & = {\mathscr{R}^\dagger \alpha_a}(\varphi) = {\mathscr{R}^\dagger \left[\tfrac{\partial}{\partial p}\mathscr{R}g_a+ \tfrac{\partial^2}{\partial p^2}\mathscr{R}f_a\right]}(\varphi) \\ & = {\mathscr{R}^\dagger \left[\tfrac{\partial}{\partial p}\hat{g}_a+ \tfrac{\partial^2}{\partial p^2}\hat{f}_a\right]}(\varphi)\\ & = {\mathscr{R}^\dagger\tfrac{\partial}{\partial p}\hat{g}_a(\varphi) + \mathscr{R}^\dagger\tfrac{\partial^2}{\partial p^2}\hat{f}_a}(\varphi) \\ &= \Big\langle\tfrac{\partial}{\partial p}\hat{g}_a,\mathscr{R}\varphi\Big\rangle + \Big\langle\tfrac{\partial^2}{\partial p^2}\hat{f}_a,\mathscr{R}\varphi\Big\rangle \\ &= \Big\langle\tfrac{\partial}{\partial p}\hat{g}_a,\hat{\varphi}\Big\rangle + \Big\langle\tfrac{\partial^2}{\partial p^2}\hat{f}_a,\hat{\varphi}\Big\rangle\\ &= \Big\langle\tfrac{\partial}{\partial p}\hat{g},\hat{\varphi}_{-a}\Big\rangle + \Big\langle\tfrac{\partial^2}{\partial p^2}\hat{f},\hat{\varphi}_{-a}\Big\rangle\\ &= - \Big\langle\hat{g},\tfrac{\partial}{\partial p}\hat{\varphi}_{-a}\Big\rangle + \Big\langle\hat{f},\tfrac{\partial^2}{\partial p^2}\hat{\varphi}_{-a}\Big\rangle\\ &= - \Big\langle g, \mathscr{R}^\dagger\tfrac{\partial}{\partial p}\hat{\varphi}_{-a}\Big\rangle + \Big\langle f,\mathscr{R}^\dagger\tfrac{\partial^2}{\partial p^2}\hat{\varphi}_{-a}\Big\rangle \end{split} $$

References

[1] Stanley Roderick Deans, The Radon transform and some of its applications, revised reprint of the 1983 original edition (English), Malabar, FL: Publishing Co., ISBN 0-89464-718-0, pp. xii+295 (1993), MR1274701, Zbl 0868.44001.

[2] Donald Ludwig, "The Radon transform on Euclidean space", (English) Communication in Pure and Applied Mathematics 19, 49-81 (1966), MR0190652, Zbl 0134.11305.

[3] Vasily Sergeevich Vladimirov (2002), Methods of the theory of generalized functions, Analytical Methods and Special Functions, Vol. 6, London–New York: Taylor & Francis, pp. XII+353, ISBN 0-415-27356-0, MR2012831, Zbl 1078.46029.